Gauss's Day of Reckoning

As with the identity of the series, the details of how Gauss solved
the problem remain a matter of conjecture. The algorithm that I
suggested—folding the sequence in half, then adding the first
and last elements, the second and next-to-last, etc.—is not
the only possibility. A related but subtly different algorithm is
mentioned by many authors. The idea is to write down the series
twice, once forward and once backward, and then add corresponding
elements. For the familiar series 1-100 this procedure yields 100
pairs of 101, for a total of 10,100; then, since the original series
was duplicated, we need to divide by 2, arriving at the correct
answer 5,050. The advantage of this scheme is that it works the same
whether the length of the sequence is odd or even, whereas the
folding algorithm requires some fussy adjustments to deal with an
odd-length series.

A third approach to the summation problem strikes me as better
still. The root idea is that for any finite set of numbers,
whether or not the numbers form an arithmetic progression, the sum
is equal to the average of all the elements multiplied by the number
of elements. Thus if you know the average, you can easily find the
sum. For most sets of numbers, this fact is not very useful, because
the only way to calculate the average is first to calculate the sum
and then divide by the number of elements. For an arithmetic
progression, however, there is a shortcut: The average over the
entire series is equal to the average of the first and last elements
(or the average of any other elements symmetrically arrayed around
the midpoint). If this was Gauss's secret weapon, then his mental
multiplication was not 50 x 101 but 100 x 50½.

All three of these ideas—and a few more besides—have
been presented by one author or another as the method that
Gauss discovered during his first arithmetic lesson. Expressed as
formulas for summing consecutive integers from 1 through n,
the three rules (folding, double rows, average) look like this:

Mathematically, it's obvious they are equivalent: For the same value
of n, they produce the same answer. But the computational
details are different and, more important, so are the reasoning
processes that lead to these formulas.

There is yet another way of thinking about the summation process:
n(n + 1)/2 has been known since antiquity as the
formula for the triangular numbers, those in the sequence 1, 3, 6,
10, 15, 21.... Thus some authors suggest that Gauss was thinking
geometrically, forming an n-by-n + 1 rectangle and
cutting it along the diagonal.